Example. For \(\mathfrak {sl}_2\), the two Clifford actions have different constituents [fcap-003F]

Take the basis \(e,f,h\) of \(\mathfrak {sl}_2(\mathbb C)\) with \[[h,e]=2e,\qquad [h,f]=-2f,\qquad [e,f]=h.\] The Killing form satisfies \(\kappa (e,f)=4\) and \(\kappa (h,h)=8\). Hence the Clifford relations are \[e^2=f^2=0,\qquad ef+fe=8,\qquad h^2=8,\qquad he=-eh,\quad hf=-fh.\] Substitution in the lift of Proposition [fcap-003E] gives \[E=\delta (e)=\frac {he}{8},\qquad F=\delta (f)=\frac {fh}{8},\qquad H=\delta (h)=\frac {ef-fe}{8}=\frac {ef}{4}-1.\] Direct Clifford multiplication yields \[E^2=F^2=0,\qquad H^2=1,\qquad EF=\frac {1+H}{2},\qquad FE=\frac {1-H}{2}.\] Thus \(E,F,EF,FE\) are the four matrix units of the lifted \(\mathfrak {sl}_2\) action.

For left multiplication, the \(H=+1\) and \(H=-1\) subspaces are interchanged by \(F\) and \(E\); the two maps are inverse on the corresponding projected spaces because \(EF\) and \(FE\) are their identity projectors. They have equal dimension, and \(\dim \mathcal {C}\kern -2pt\ell (Q_\kappa )=8\), so each has dimension four. Choosing a basis of the positive space and its \(F\)-images displays \[(\mathcal {C}\kern -2pt\ell (Q_\kappa ),L)\simeq 4V_1,\] four copies of the standard two-dimensional module.

For the commutator action \(D\), antisymmetrization identifies the same vector space with \(\bigwedge \mathfrak {sl}_2\). The exterior powers are \[\bigwedge ^0\mathfrak g\simeq V_0,\qquad \bigwedge ^1\mathfrak g\simeq V_2,\qquad \bigwedge ^2\mathfrak g\simeq V_2,\qquad \bigwedge ^3\mathfrak g\simeq V_0,\] where the middle identification uses the bracket and the last uses the invariant volume. Therefore \[(\mathcal {C}\kern -2pt\ell (Q_\kappa ),D)\simeq 2V_0\oplus 2V_2.\] The dimensions agree but the characters do not. [kostant1997clifford, Theorem 39, pp. 312--313] gives the first answer through Kostant's primary-type theorem. [panyushev2001exterior, Example 2.5(1), internal preprint p. 8] independently gives the exterior decomposition. The displayed matrix-unit calculation directly verifies which action produces which answer.